Easy
For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the equilibrium constant is $64$. If the volume of the container is reduced to $1/4$ of its original value,the value of the equilibrium constant will be .......
- A$16$
- B$32$
- C$64$
- D$128$
Explore More
Similar Questions
Chemical equilibrium is dynamic in nature because
Easy
View SolutionExplain the dynamic nature of chemical equilibrium with a suitable reaction example.
Difficult
View SolutionIn which of the following reactions,the concentration of product is higher than the concentration of reactant at equilibrium? $(K = \text{equilibrium constant})$
For a reversible reaction involving two reactants,if the concentration of the reactants is doubled,the equilibrium constant will:
Medium
View Solution$A$ reversible reaction $H_2 + Cl_2 \rightleftharpoons 2HCl$ is carried out in a $1 \ L$ flask. If the same reaction is carried out in a $2 \ L$ flask,the equilibrium constant will be
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo